A 0.125 kg hockey puck moving at 20.0 m/s is caught and held by an 85.0 kg goalkeeper at rest. After catching the puck, with wha
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Missing question in the text of the exercise. Found on internet:
"What is the acceleration due to gravity on the surface of the planet?"
Solution:
The gravitational acceleration at Earth's surface is given by:
where
G is the gravitational constant
M is the Earth's mass
r is the Earth's radius
The Earth's mass can be rewritten also as the product between the Earth's density,
, and its volume (the volume of a sphere of radius r):
Now let's call M' the mass of the new planet, r' its radius and d' its density. The acceleration due to gravity on the surface of the new planet is
(1)
so we need to find M' and r'.
The problem says the radius of the new planet is twice the Earth's radius:
(2)
and that its density is 2/3 of Earth's density:
so the mass M' of the new planet is, with respect to the Earth's mass:
(3)
And if we substitute (2) and (3) into (1), we find the gravitational acceleration on the surface of the new planet:
And since
, we find
"What is the acceleration due to gravity on the surface of the planet?"
Solution:
The gravitational acceleration at Earth's surface is given by:
where
G is the gravitational constant
M is the Earth's mass
r is the Earth's radius
The Earth's mass can be rewritten also as the product between the Earth's density,
Now let's call M' the mass of the new planet, r' its radius and d' its density. The acceleration due to gravity on the surface of the new planet is
so we need to find M' and r'.
The problem says the radius of the new planet is twice the Earth's radius:
and that its density is 2/3 of Earth's density:
so the mass M' of the new planet is, with respect to the Earth's mass:
And if we substitute (2) and (3) into (1), we find the gravitational acceleration on the surface of the new planet:
And since
Answer:
m = 3,265 10⁻²⁰ E
Explanation:
For this exercise we can use Newton's second law applied to our system, which consists of a capacitor that creates the uniform electric field and the drop of oil with two extra electrons.
∑ F = 0
- W = 0
the electric force is
F_{e} = q E
as they indicate that the charge is two electrons
F_{e} = 2e E
The weight is given by the relationship
W = mg
we substitute in the first equation
2e E = m g
m = 2e E / g
let's put the value of the constants
m = (2 1.6 10⁻¹⁹ / 9.80) E
m = 3,265 10⁻²⁰ E
The value of the electric field if it is a theoretical problem must be given and if it is an experiment it can be calculated with measures of the spacing between plates and the applied voltage, so that the system is in equilibrium